 |
| ||||
Home Page
F.A.Qs
Statistical Charts
Past Contests
Scheduled Contests
Award Contest
| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
Re:拓扑排序后 dp[a][a]=1 dp[a][b]=dp[b][a]=sum(dp[a][parent[b]])/2 when order(a)>order(b)In Reply To:拓扑排序后 dp[a][a]=1 dp[a][b]=dp[b][a]=sum(dp[a][parent[b]]) when order(a)>order(b) Posted by:lddlinan at 2020-06-30 19:23:48 Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator
